In the previous section, we saw power derived from a torque. In this section, we will see .angular momentum in case of rotation about a fixed axis'
In chap 7.17, we saw the basics about angular momentum of a system of particles
We saw the following information:
Information 1:
Eq.7.19: dldt=→r×→F=→τ
• That is: 'Rate of change of angular momentum' calculated over an instant is the torque experienced by the particle at that instant
Information 2:
• To get the total angular momentum →L of a system of particles, we add the angular momenta of individual particles
• Thus for a system of n particles, we have:
→L=→l1+→l1+→l1+...+→ln=i=n∑i=1→li
• Note that, this is an addition of vectors. That is., we have to add the momenta vectorially
• In the above equation, we know that: →li=→ri×→pi
• So the equation in (1) becomes:
Eq.7.20: →L=i=n∑i=1(→ri×→pi)
• If required, the →pi can be further split up as: →pi=mi×→vi
Information 3:
Eq.7.21: dLdt=→τext
• Based on this equation, we can write:
The 'rate of change of angular momentum of a system of particles' at any instant is equal to the 'sum of external torques on all the particles'
In chap 7.17, we saw the basics about angular momentum of a system of particles
We saw the following information:
Information 1:
Eq.7.19: dldt=→r×→F=→τ
• That is: 'Rate of change of angular momentum' calculated over an instant is the torque experienced by the particle at that instant
Information 2:
• To get the total angular momentum →L of a system of particles, we add the angular momenta of individual particles
• Thus for a system of n particles, we have:
→L=→l1+→l1+→l1+...+→ln=i=n∑i=1→li
• Note that, this is an addition of vectors. That is., we have to add the momenta vectorially
• In the above equation, we know that: →li=→ri×→pi
• So the equation in (1) becomes:
Eq.7.20: →L=i=n∑i=1(→ri×→pi)
• If required, the →pi can be further split up as: →pi=mi×→vi
Information 3:
Eq.7.21: dLdt=→τext
• Based on this equation, we can write:
The 'rate of change of angular momentum of a system of particles' at any instant is equal to the 'sum of external torques on all the particles'
We now want to see a special case:
The angular momentum about a fixed axis
We will write it in steps:
1. From information 2 above, we have:
→L=i=n∑i=1(→ri×→pi)
2. In this equation, (→ri×→pi) is the angular momentum of a single particle (the ith particle)
• Also recall that the angular momentum of a single particle is denoted by '→l'
• So the angular momentum of the ith particle is denoted by '→l'
• That is., →li=(→ri×→pi)
3. We will consider that typical particle first.
• That is., we will find the '→li of a typical particle about a fixed axis' first
• Then we will sum up the angular momenta of all the particles to get the →L of the whole body
4. Consider again fig.7.79 that we saw in a previous section. It is shown again below:
• For the single particle (indicated by the red sphere), we have: →l=→r×→p
5. →r can be split up as: →r=→OC+→CP
• So we get: →l=(→OC+→CP)×→p
⇒→l=(→OC×→p+→CP×→p)
• But →p=m×→v
• So we get: →l=[→OC×(m→v)+→CP×(m→v)]
6. There are two terms on the right side
• Consider the first term: →OC×(m→v)
• It is the cross product of two vectors: →OC and (m→v)
• Obviously the resulting vector will be perpendicular to →OC
• The vector →OC is along the axis of rotation
• So the resulting vector from the first term is perpendicular to the axis of rotation
7. Consider the second term: →CP×(m→v)
• →CP lies on the plane of the red circle
• (m→v) also lies on the plane of the red circle
• So the resulting vector will be perpendicular to the plane of the red circle
• That means, the resulting vector will be parallel to the axis of rotation
8. Let us calculate this 'parallel vector' obtained from the second term: →CP×(m→v)
• We will apply the cross product rule: |(→A×→B)|=|→A|×|→B|×sinθ
• In the present case, we have:
♦ |→CP|=r⊥
♦ |→v|=r⊥|→ω|
♦ Angle between the two vectors = 90o. So sin θ = sin 90 = 1
• Thus we get: |(→CP×(m→v))|=r⊥×(mr⊥|→ω|)×1
⇒|(→CP×(m→v))|=(mr2⊥|→ω|)
9. This is the magnitude of the resulting vector of the second term
• We want the direction also
• We saw that, the resulting vector is parallel to the axis of rotation
♦ The axis of rotation is the z-axis
♦ The unit vector along the z-axis is ˆk
• So we get:
The resulting vector from the second term is: (mr2⊥|→ω|)ˆk
10. Since this vector is parallel to the axis of rotation (the z-axis), we will denote it as →lz
• So we can write: →lz=(mr2⊥|→ω|)ˆk
• So the result in (5) becomes: →l=→lz+(→OC×(m→v))
11. This is a vector sum
• The first component →lz is parallel to the axis of rotation
• The second component (→OC×(m→v)) is not parallel to the axis of rotation
• So the vector sum →l=→lz+(→OC×(m→v)) is not parallel to the axis of rotation
12. We know that the angular velocity →ω is parallel to the axis of rotation (See fig.7.78 in chapter 7.15)
• So we get an important result:
The angular velocity →ω and angular momentum →l of a particle are not necessarily parallel
• A comparison with translational motion:
In translational motion we have:
Linear velocity →v and linear momentum →p are always parallel to each other
• Now we add the angular momenta of all the particles
• The sum will give the angular momentum →L of the whole body
14. So we can write:
L=i=n∑i=1(li)
⇒L=i=n∑i=1(→lz(i))+i=n∑i=1(→OC(i)×(m(i)→v(i)))
Where:
• →lz(i) is the 'angular momentum component' of the ith parallel to the axis of rotation
• →OC(i) is the vector between the two points:
♦ Origin O of the reference system
♦ Center C of the circle described by the ith particle
• m(i) is the mass of the ith particle
• →v(i) is the linear velocity of the ith particle
15. Thus we see that, L also has two components
Consider the first component: i=n∑i=1(→lz(i))
• All vectors involved in this summation are parallel to the z-axis (axis of rotation)
• So the result of the summation will be parallel to the z-axis
• So we will denote this component as →Lz
16. Consider the second component: i=n∑i=1(→OC(i)×(m(i)→v(i)))
• All vectors involved in this summation are perpendicular to the z-axis. We know the reason:
♦ The vector product →OC(i)×(m(i)→v(i)) will be perpendicular to →OC(i)
♦ →OC(i) lies along the z-axis
• So the result of the summation will be perpendicular to the z-axis
• So we will denote this component as →L⊥
17. Thus we can write:
Eq.7.30:
→L=→Lz+→L⊥
• Let us consider each component separately:
We have: →Lz=i=n∑i=1(→lz(i))
• But from (10), we have: →lz(i)=(m(i)r2⊥(i)|→ω|)ˆk
• So, when we take the summation i=n∑i=1(→lz(i)),
We are actually taking the summation: i=n∑i=1(m(i)r2⊥(i)|→ω|)ˆk
• |→ω|ˆk is the same for all particles. So it can be taken outside
• So we get: i=n∑i=1(→lz(i))=|→ω|ˆki=n∑i=1(m(i)r2⊥(i))
18. But i=n∑i=1(m(i)r2⊥(i)) = I, the moment of inertia of the whole body (Eq.7.25 in chapter 7.23)
So we get: →Lz=i=n∑i=1(→lz(i))=I|→ω|ˆk
19. Now consider the second component: →L⊥=i=n∑i=1(→OC(i)×(m(i)→v(i)))
• This is a summation of vectors
• Each of those vectors results from the 'vector multiplication' of two vectors: →OC(i)and(m(i)→v(i))
Let us see an example:
(i) Consider the square rod shown in fig.7.136 (a) below:
• It is rotating about the axis shown in blue color
• The direction of rotation is indicated by the yellow curved arrow
• It is a uniform square rod. Also, the blue axis passes through the exact center of the rod
(ii) A particle is isolated at position 'P' along the rod
• It is shown in red color
• Let us write the properties of that particle:
♦ →OC(i)=→OC
♦ m(i)=mP
♦ →v(i)=→v(P)
• So for the particle at P, we can easily calculate the cross product of →OC(i)and(m(i)→v(i))
(iii) For the particle at P, there will be an exact replica on the other side of the axis
• This particle is at Q
(An exact replica is obtained because, the square rod is uniform and also, the axis passes through the exact center)
• The position of Q will be such that, CQ = CP
• Let us write the properties of the particle at Q:
♦ →OC(i)=→OC
♦ m(i)=mQ
♦ →v(i)=→v(Q)
• So for the particle at Q also, we can easily calculate the cross product of →OC(i)and(m(i)→v(i))
(iv) Let us write a comparison of properties:
• For both the particles, →OC(i)=→OC
• Because of the symmetry, m(P)=mQ
• Because of the symmetry, the magnitudes of →v(P)and→v(Q) will be the same
♦ But their directions will be exactly opposite to each other
(v) Consider the two cross products:
(a) Cross product of →OC(P)and(m(P)→v(P))
(b) Cross product of →OC(Q)and(m(Q)→v(Q))
• The 'vector obtained as the cross product in (a)' will be equal and opposite to the 'vector obtained as the cross product in (b)'
• So in the summation, the two vectors will cancel each other
(vi) For each particle in the rod, there will be an exact replica on the other side of the axis
• So all particles in the rod, can be grouped into pairs such that:
• In any pair, one member is the exact replica of the other
(vii) If we draw a circle with center at C and passing through P,
♦ The point Q will also lie on the circle
♦ The point Q qill be diametrically opposite to P
• This is true for all symmetric bodies
• The members of the pairs will lie diametrically opposite on the circle with center at C
(vii) So the net result is that, the summation →L⊥=i=n∑i=1(→OC(i)×(m(i)→v(i))) will become zero
• That means →L⊥=0
20. So the result in (1) becomes: →L=→Lz
• But for this simplification, two conditions should be satisfied:
(i) The object must be uniform and symmetric
(ii) The rotation must be about the axis of symmetry
• For our present discussion we consider only those rotations which satisfy the two conditions
• So we can confidently use the simplified form
We can write:
Eq.7.31:
For symmetric rotation of symmetric bodies, →L=I|→ω|ˆk
The angular momentum about a fixed axis
We will write it in steps:
1. From information 2 above, we have:
→L=i=n∑i=1(→ri×→pi)
2. In this equation, (→ri×→pi) is the angular momentum of a single particle (the ith particle)
• Also recall that the angular momentum of a single particle is denoted by '→l'
• So the angular momentum of the ith particle is denoted by '→l'
• That is., →li=(→ri×→pi)
3. We will consider that typical particle first.
• That is., we will find the '→li of a typical particle about a fixed axis' first
• Then we will sum up the angular momenta of all the particles to get the →L of the whole body
4. Consider again fig.7.79 that we saw in a previous section. It is shown again below:
 |
| Fig.7.79 |
5. →r can be split up as: →r=→OC+→CP
• So we get: →l=(→OC+→CP)×→p
⇒→l=(→OC×→p+→CP×→p)
• But →p=m×→v
• So we get: →l=[→OC×(m→v)+→CP×(m→v)]
6. There are two terms on the right side
• Consider the first term: →OC×(m→v)
• It is the cross product of two vectors: →OC and (m→v)
• Obviously the resulting vector will be perpendicular to →OC
• The vector →OC is along the axis of rotation
• So the resulting vector from the first term is perpendicular to the axis of rotation
7. Consider the second term: →CP×(m→v)
• →CP lies on the plane of the red circle
• (m→v) also lies on the plane of the red circle
• So the resulting vector will be perpendicular to the plane of the red circle
• That means, the resulting vector will be parallel to the axis of rotation
8. Let us calculate this 'parallel vector' obtained from the second term: →CP×(m→v)
• We will apply the cross product rule: |(→A×→B)|=|→A|×|→B|×sinθ
• In the present case, we have:
♦ |→CP|=r⊥
♦ |→v|=r⊥|→ω|
♦ Angle between the two vectors = 90o. So sin θ = sin 90 = 1
• Thus we get: |(→CP×(m→v))|=r⊥×(mr⊥|→ω|)×1
⇒|(→CP×(m→v))|=(mr2⊥|→ω|)
9. This is the magnitude of the resulting vector of the second term
• We want the direction also
• We saw that, the resulting vector is parallel to the axis of rotation
♦ The axis of rotation is the z-axis
♦ The unit vector along the z-axis is ˆk
• So we get:
The resulting vector from the second term is: (mr2⊥|→ω|)ˆk
10. Since this vector is parallel to the axis of rotation (the z-axis), we will denote it as →lz
• So we can write: →lz=(mr2⊥|→ω|)ˆk
• So the result in (5) becomes: →l=→lz+(→OC×(m→v))
11. This is a vector sum
• The first component →lz is parallel to the axis of rotation
• The second component (→OC×(m→v)) is not parallel to the axis of rotation
• So the vector sum →l=→lz+(→OC×(m→v)) is not parallel to the axis of rotation
12. We know that the angular velocity →ω is parallel to the axis of rotation (See fig.7.78 in chapter 7.15)
• So we get an important result:
The angular velocity →ω and angular momentum →l of a particle are not necessarily parallel
• A comparison with translational motion:
In translational motion we have:
Linear velocity →v and linear momentum →p are always parallel to each other
13. So we calculated the angular momentum of a single particle
• It is given by the expression in (10)
• It is given by the expression in (10)
• The sum will give the angular momentum →L of the whole body
14. So we can write:
L=i=n∑i=1(li)
⇒L=i=n∑i=1(→lz(i))+i=n∑i=1(→OC(i)×(m(i)→v(i)))
Where:
• →lz(i) is the 'angular momentum component' of the ith parallel to the axis of rotation
• →OC(i) is the vector between the two points:
♦ Origin O of the reference system
♦ Center C of the circle described by the ith particle
• m(i) is the mass of the ith particle
• →v(i) is the linear velocity of the ith particle
15. Thus we see that, L also has two components
Consider the first component: i=n∑i=1(→lz(i))
• All vectors involved in this summation are parallel to the z-axis (axis of rotation)
• So the result of the summation will be parallel to the z-axis
• So we will denote this component as →Lz
16. Consider the second component: i=n∑i=1(→OC(i)×(m(i)→v(i)))
• All vectors involved in this summation are perpendicular to the z-axis. We know the reason:
♦ The vector product →OC(i)×(m(i)→v(i)) will be perpendicular to →OC(i)
♦ →OC(i) lies along the z-axis
• So the result of the summation will be perpendicular to the z-axis
• So we will denote this component as →L⊥
17. Thus we can write:
Eq.7.30:
→L=→Lz+→L⊥
• Let us consider each component separately:
We have: →Lz=i=n∑i=1(→lz(i))
• But from (10), we have: →lz(i)=(m(i)r2⊥(i)|→ω|)ˆk
• So, when we take the summation i=n∑i=1(→lz(i)),
We are actually taking the summation: i=n∑i=1(m(i)r2⊥(i)|→ω|)ˆk
• |→ω|ˆk is the same for all particles. So it can be taken outside
• So we get: i=n∑i=1(→lz(i))=|→ω|ˆki=n∑i=1(m(i)r2⊥(i))
18. But i=n∑i=1(m(i)r2⊥(i)) = I, the moment of inertia of the whole body (Eq.7.25 in chapter 7.23)
So we get: →Lz=i=n∑i=1(→lz(i))=I|→ω|ˆk
19. Now consider the second component: →L⊥=i=n∑i=1(→OC(i)×(m(i)→v(i)))
• This is a summation of vectors
• Each of those vectors results from the 'vector multiplication' of two vectors: →OC(i)and(m(i)→v(i))
Let us see an example:
(i) Consider the square rod shown in fig.7.136 (a) below:
 |
| Fig.7.136 |
• The direction of rotation is indicated by the yellow curved arrow
• It is a uniform square rod. Also, the blue axis passes through the exact center of the rod
(ii) A particle is isolated at position 'P' along the rod
• It is shown in red color
• Let us write the properties of that particle:
♦ →OC(i)=→OC
♦ m(i)=mP
♦ →v(i)=→v(P)
• So for the particle at P, we can easily calculate the cross product of →OC(i)and(m(i)→v(i))
(iii) For the particle at P, there will be an exact replica on the other side of the axis
• This particle is at Q
(An exact replica is obtained because, the square rod is uniform and also, the axis passes through the exact center)
• The position of Q will be such that, CQ = CP
• Let us write the properties of the particle at Q:
♦ →OC(i)=→OC
♦ m(i)=mQ
♦ →v(i)=→v(Q)
• So for the particle at Q also, we can easily calculate the cross product of →OC(i)and(m(i)→v(i))
(iv) Let us write a comparison of properties:
• For both the particles, →OC(i)=→OC
• Because of the symmetry, m(P)=mQ
• Because of the symmetry, the magnitudes of →v(P)and→v(Q) will be the same
♦ But their directions will be exactly opposite to each other
(v) Consider the two cross products:
(a) Cross product of →OC(P)and(m(P)→v(P))
(b) Cross product of →OC(Q)and(m(Q)→v(Q))
• The 'vector obtained as the cross product in (a)' will be equal and opposite to the 'vector obtained as the cross product in (b)'
• So in the summation, the two vectors will cancel each other
(vi) For each particle in the rod, there will be an exact replica on the other side of the axis
• So all particles in the rod, can be grouped into pairs such that:
• In any pair, one member is the exact replica of the other
(vii) If we draw a circle with center at C and passing through P,
♦ The point Q will also lie on the circle
♦ The point Q qill be diametrically opposite to P
• This is true for all symmetric bodies
• The members of the pairs will lie diametrically opposite on the circle with center at C
(vii) So the net result is that, the summation →L⊥=i=n∑i=1(→OC(i)×(m(i)→v(i))) will become zero
• That means →L⊥=0
20. So the result in (1) becomes: →L=→Lz
• But for this simplification, two conditions should be satisfied:
(i) The object must be uniform and symmetric
(ii) The rotation must be about the axis of symmetry
• For our present discussion we consider only those rotations which satisfy the two conditions
• So we can confidently use the simplified form
We can write:
Eq.7.31:
For symmetric rotation of symmetric bodies, →L=I|→ω|ˆk
Let us write a summary about the above discussion:
■ In chap 7.17, we obtained the general equation for the angular momentum of any object:
Eq.7.20: →L=i=n∑i=1(→ri×→pi)
■ In the present section, we applied it to objects which rotate about a fixed axis. We obtained:
Eq.7.30:
→L=→Lz+→L⊥
■ Also, we applied it to symmetric objects which rotate symmetrically about a fixed axis. We obtained:
Eq.7.31:
For symmetric rotation of symmetric bodies, →L=I|→ω|ˆk
■ In chap 7.17, we obtained the general equation for the angular momentum of any object:
Eq.7.20: →L=i=n∑i=1(→ri×→pi)
■ In the present section, we applied it to objects which rotate about a fixed axis. We obtained:
Eq.7.30:
→L=→Lz+→L⊥
■ Also, we applied it to symmetric objects which rotate symmetrically about a fixed axis. We obtained:
Eq.7.31:
For symmetric rotation of symmetric bodies, →L=I|→ω|ˆk
In the next section, we will see applications of Eq.7.30
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